PART A The element Si belongs to IVA group. Therefore, four electro.pdf
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PART A: The element Si belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D: All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons. PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is [Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal. PART I: The element Ba belongs to IIA group. Therefore, two electrons present in the s-orbital. Hence the condensed electron configuration of Ba is [Xe] 6s2. PART J: The outermost s- electron is completely filled. Therefore, there are no unpaired electrons present in Ba metal. PART K: The element Tl belongs to IIIA group. Therefore, the condensed electron configuration of Tl is [Xe] 4f14 5d10 6s2 6p1. PART L: The all orbitals are filled with electrons except Tl. Therefore, the number of unpaired electrons is 1 Solution PART A: The element Si belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D: All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons. PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is [Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal. PART I: The element Ba belongs to IIA group. Th.
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![PART A: The element Si belongs to IVA group. Therefore, four electrons present in the
outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The
s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the
number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs
to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D:
All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons.
PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present
after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is
[Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons
present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The
element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell.
Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital
and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal.
PART I: The element Ba belongs to IIA group. Therefore, two electrons present in the s-orbital.
Hence the condensed electron configuration of Ba is [Xe] 6s2. PART J: The outermost s-
electron is completely filled. Therefore, there are no unpaired electrons present in Ba metal.
PART K: The element Tl belongs to IIIA group. Therefore, the condensed electron configuration
of Tl is [Xe] 4f14 5d10 6s2 6p1. PART L: The all orbitals are filled with electrons except Tl.
Therefore, the number of unpaired electrons is 1
Solution
PART A: The element Si belongs to IVA group. Therefore, four electrons present in the
outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The
s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the
number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs
to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D:
All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons.
PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present
after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is
[Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons
present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The
element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell.
Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital
and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal.
PART I: The element Ba belongs to IIA group. Therefore, two electrons present in the s-orbital.
Hence the condensed electron configuration of Ba is [Xe] 6s2. PART J: The outermost s-](https://image.slidesharecdn.com/partatheelementsibelongstoivagroup-230628084039-e5fa1c82/85/PART-A-The-element-Si-belongs-to-IVA-group-Therefore-four-electro-pdf-1-320.jpg)
![electron is completely filled. Therefore, there are no unpaired electrons present in Ba metal.
PART K: The element Tl belongs to IIIA group. Therefore, the condensed electron configuration
of Tl is [Xe] 4f14 5d10 6s2 6p1. PART L: The all orbitals are filled with electrons except Tl.
Therefore, the number of unpaired electrons is 1](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Be careful; nitrogen is NOT a transition element. The element you ar.pdf
Be careful; nitrogen is NOT a transition element. The element you are thinking of is nickel,
which is a transition element and has the chemical symbol \"Ni\" (different from that of nitrogen,
which is \"N\"). Now, transition elements are those elements whose atoms have incomplete d-
sub-shells, meaning that their electron configurations include a partly, not completely, filled d-
sub-shell.
(1) Let\'s take a look at the electron configuration of nickel:
[Ar] 4s2 3d8
Notice that the d subshell, only contains 8 electrons, while elements whose d sub-shells are full
have 10 electrons. Also, sorry, but nitrogen can\'t possibly be a transition element because it
doesn\'t have any d electrons in its electron configuration,
(2) Why are such special elements called \"transition elements\"?
Transition elements are placed on the periodic table in the intermediate d-block, which is the
relatively large middle section of the table which partitions alkali metals and alkaline earth
metals to the left of the table, and metalloids and non-metal elements to the right of the table.
Also, the physiochemical properties of transition elements are similarly \"in transition\"; the
unpaired, low-reactivity d electrons cause these elements to have several different oxidation
states, thus causing them to be in a state of transition.
1s2 2s2 2p3
Solution
Be careful; nitrogen is NOT a transition element. The element you are thinking of is nickel,
which is a transition element and has the chemical symbol \"Ni\" (different from that of nitrogen,
which is \"N\"). Now, transition elements are those elements whose atoms have incomplete d-
sub-shells, meaning that their electron configurations include a partly, not completely, filled d-
sub-shell.
(1) Let\'s take a look at the electron configuration of nickel:
[Ar] 4s2 3d8
Notice that the d subshell, only contains 8 electrons, while elements whose d sub-shells are full
have 10 electrons. Also, sorry, but nitrogen can\'t possibly be a transition element because it
doesn\'t have any d electrons in its electron configuration,
(2) Why are such special elements called \"transition elements\"?
Transition elements are placed on the periodic table in the intermediate d-block, which is the
relatively large middle section of the table which partitions alkali metals and alkaline earth
metals to the left of the table, and metalloids and non-metal elements to the right of the table.
Also, the physiochemical properties of transition elements are similarly \"in transition\"; the
unpaired, low-reactivity d electrons cause these elements to have several different oxidation
states, thus causing them to be in a state of transition.
1s2 2s2 2p3.
Recommended
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Chapter7
First Year in Dentistry First Semester
Al-Azhar University - Gaza
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What is a transition element
Transition elements A LEVEL CIE CHEMISTRY
presentation questions success a level A Star 9
Be careful; nitrogen is NOT a transition element. The element you ar.pdf
Be careful; nitrogen is NOT a transition element. The element you are thinking of is nickel,
which is a transition element and has the chemical symbol \"Ni\" (different from that of nitrogen,
which is \"N\"). Now, transition elements are those elements whose atoms have incomplete d-
sub-shells, meaning that their electron configurations include a partly, not completely, filled d-
sub-shell.
(1) Let\'s take a look at the electron configuration of nickel:
[Ar] 4s2 3d8
Notice that the d subshell, only contains 8 electrons, while elements whose d sub-shells are full
have 10 electrons. Also, sorry, but nitrogen can\'t possibly be a transition element because it
doesn\'t have any d electrons in its electron configuration,
(2) Why are such special elements called \"transition elements\"?
Transition elements are placed on the periodic table in the intermediate d-block, which is the
relatively large middle section of the table which partitions alkali metals and alkaline earth
metals to the left of the table, and metalloids and non-metal elements to the right of the table.
Also, the physiochemical properties of transition elements are similarly \"in transition\"; the
unpaired, low-reactivity d electrons cause these elements to have several different oxidation
states, thus causing them to be in a state of transition.
1s2 2s2 2p3
Solution
Be careful; nitrogen is NOT a transition element. The element you are thinking of is nickel,
which is a transition element and has the chemical symbol \"Ni\" (different from that of nitrogen,
which is \"N\"). Now, transition elements are those elements whose atoms have incomplete d-
sub-shells, meaning that their electron configurations include a partly, not completely, filled d-
sub-shell.
(1) Let\'s take a look at the electron configuration of nickel:
[Ar] 4s2 3d8
Notice that the d subshell, only contains 8 electrons, while elements whose d sub-shells are full
have 10 electrons. Also, sorry, but nitrogen can\'t possibly be a transition element because it
doesn\'t have any d electrons in its electron configuration,
(2) Why are such special elements called \"transition elements\"?
Transition elements are placed on the periodic table in the intermediate d-block, which is the
relatively large middle section of the table which partitions alkali metals and alkaline earth
metals to the left of the table, and metalloids and non-metal elements to the right of the table.
Also, the physiochemical properties of transition elements are similarly \"in transition\"; the
unpaired, low-reactivity d electrons cause these elements to have several different oxidation
states, thus causing them to be in a state of transition.
1s2 2s2 2p3.
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This part contains ionisation enthalpies, oxidation states, metal oxides & oxocations, magnetic properties, coloured ions of d block elements, catalysts, interstitial compounds, alloy formation & some important conceptual questions with answers as hints. Also some reasoning questions are given to test the understanding of properties of d block elements.
Answer Every investor expects dividend from his investments.Dividen.pdf
Answer: Every investor expects dividend from his investments.Dividend income helps investors
in dealing with their routine expenses. Investors put their money to earn some positive returns
and dividend is one of the way of getting returns for investors. So, It is a duty of management to
satisfy dividend expectation of investors.
If management do not pay dividend then it is known as retention of earnings. Retention earnings
has it\'s own pro and cons, which are described as follows:
Pros:
Retained earnings reduces organization\'s dependence on external borrowings. It can be
considered as an advantage when external borrowings are not available easily.
Retained earnings indicates good business opportunities, It means organization is growing
financially. It may result in better stock valuation of organization.
In some situation dividend is subject to taxes (either distribution tax or income tax) in the hands
of investors. So, in this case investors are not interested in dividend but they are more interested
in capital gain(which may be tax exempt). So, In this situation retained earnings fulfill investor\'s
objective.
Cons:
As explained earlier all the investors have some expectation of dividend, in case of higher
retention of earnings it may become difficult to fulfill investor\'s expectations.
Retention of earnings may affect investor\'s earning capacity when investors have available
opportunities, which gives higher return than organization earns by retaining amount. In this case
investor looses earning opportunities.
In case of dividend management\'s decision is significantly influenced by media and institutional
investors. Media is responsible for organization\'s image in society, such image significantly
affects organization\'s financial performance. This is why management always considers media\'s
point of view before making any dividend decision.
Institutional investors are usually invests their money and takes part in decision making process
of organization. Institutional investor\'s weight in decision making process is generally very
high, so it is obvious that they influence management\'s dividend decision also.
Solution
Answer: Every investor expects dividend from his investments.Dividend income helps investors
in dealing with their routine expenses. Investors put their money to earn some positive returns
and dividend is one of the way of getting returns for investors. So, It is a duty of management to
satisfy dividend expectation of investors.
If management do not pay dividend then it is known as retention of earnings. Retention earnings
has it\'s own pro and cons, which are described as follows:
Pros:
Retained earnings reduces organization\'s dependence on external borrowings. It can be
considered as an advantage when external borrowings are not available easily.
Retained earnings indicates good business opportunities, It means organization is growing
financially. It may result in better stock valuation of organization.
In some situ.
Ans. Gene is defined as the segment of DNA that gives a functional p.pdf
Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A
functional product can be either protein (or polypeptide) or RNA.
mRNA transcribes from a gene is further translated to produce polypeptides. However, all other
RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and
various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid
carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA
mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene
regulations.
SO, polypeptides are just one class of functional product from DNA. The genomic DNA also
produces various classes of RNAs with diverse functions that are crucial for cellular activates
and gene regulation.
Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides
can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all
genes (including those encoding polypeptides and RNAs).
Solution
Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A
functional product can be either protein (or polypeptide) or RNA.
mRNA transcribes from a gene is further translated to produce polypeptides. However, all other
RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and
various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid
carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA
mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene
regulations.
SO, polypeptides are just one class of functional product from DNA. The genomic DNA also
produces various classes of RNAs with diverse functions that are crucial for cellular activates
and gene regulation.
Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides
can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all
genes (including those encoding polypeptides and RNAs)..
additional optmization techniques for underlying IP network must1.pdf
additional optmization techniques for underlying IP network must:
1.Have bandwidth and performance to handle converged services.
2.Seprate VLAN for Voice traffic.
3.Meet the demand of high availability of voice services by providing resiliency to mitigate the
effect of network outages.
4.Should be modular, hierarchical & consistent to promote consistency and manageability.
5.configure QoS on core switch.
Solution
additional optmization techniques for underlying IP network must:
1.Have bandwidth and performance to handle converged services.
2.Seprate VLAN for Voice traffic.
3.Meet the demand of high availability of voice services by providing resiliency to mitigate the
effect of network outages.
4.Should be modular, hierarchical & consistent to promote consistency and manageability.
5.configure QoS on core switch..
According to the given equation, aqueous carbon dioxide reacts with .pdf
According to the given equation, aqueous carbon dioxide reacts with water to form carbonic
acid, Carbonic acid is not stable in normal body temperatures, therefore, it ionizes into
bicarbonate ion (HCO3-) and a hydrogen ion (H+) is liberated. The liberation of hydrogen ions
decreases the pH.
During breathing, oxygen is inhaled which is carried by the blood to the different tissues of the
body. Oxygen facilitates the various metabolic reactions occurring inside our cells and helps in
respiration, or the generation of energy. Carbon dioxide, formed as a by-product of these energy-
generating metabolic reactions, diffuses out of the tissues and into the blood stream, which
carries CO2 to the lungs, where it is exhaled out.
a) In case of hypoventilation :
The exchange of gases that occurs in the lungs while breathing is impaired. Therefore, the
requisite amount of carbon dioxide is not being exhaled out. This leads to a build-up of carbon-
dioxide in the blood plasma. As a result, the forward reaction is facilitated and more H+ ions are
generated. The higher concentration of H+ ions decreases the pH (acidic pH). This can lead to
acidosis.
The build-up of carbon dioxide in the blood facilitates the forward reaction and therefore the
equilibrium shifts to the right.
The hydronium ion or H3O+ is the product of a hydrogen ion (H+) and water. Since a hydrogen
ion is too small to exist freely in water, it therefore combines with a water molecule to form a
hydronium ion which is more stable. In case of hypoventilation, the concentration of H+ ions
increase. Therefore, the concentration of hydronium ions increase during hypoventilation.
The stress concentration or partial pressure of CO2 during hypoventilation is significantly higher
than 40mmHg (around 60-65 mmHg).
b) In case of hyperventilation :
The breathing rate is more than normal. Therefore, as a result of heavier and faster breathing, the
CO2 concentration in the blood rapidly decreases, that is, CO2 easily leaves the blood plasma.
Decrease in the CO2 concentration slows down the rate of the forward reaction, reducing the
levels at which bicarbonate ion and H+ ion concentration were being generated. The bicarbonate
and hydrogen ions recombine to yield CO2 and water. This lowering in hydrogen ion
concentration consequently lowers the pH (alkaline pH). This can lead to alkalosis.
As a result of the lowering of CO2 , the reverse reaction takes place. The carbonic acid changes
back to CO2 and H2O, shifting the equilibrium to the left.
The concentration of H3O+ decreases during hyperventilation, since the H+ ion concentration is
lowered.
The stress levels or partial pressure of CO2 during hyperventilation is significantly lower than 40
mmHg (around 25-30 mmHg).
c) If sodium carbonate or Na2CO3 is added instead of sodium bicarbonate, then sodium
carbonate would react with water and it would hydrolyse into Na+ and CO32- ions. The CO32-
combine with the hydrogen (H) present in water to form H2CO3 (carbon.
a) mean = 1.43Thus distribution is Poisson(4.2)P(X = 4) = 4.2^4.pdf
a) mean = 1.4*3
Thus distribution is Poisson(4.2)
P(X = 4) = 4.2^4 * e^{-4.2} / 4! = 0.1944
b)
average number of ghosts in 90min (1.5hrs) interval = 1.4*1.5 = 2.1
c)
Probability that 4th ghost will appear before 10 am
= Probability that number of ghosts appearing from 8 to 10 am >= 4
As it is a 2 hr period, the distribution of number of ghosts is Poisson(1.4*2) i.e Poisson(2.8)
P(X >= 4)
= 1 - P(X < 4)
= 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]
= 1 - [(2.8^0 * e^{-2.8} / 0!) + (2.8^1 * e^{-2.8} / 1!) + (2.8^2 * e^{-2.8} / 2!) + (2.8^3 * e^{-
2.8} / 3!)]
= 1 - [0.0608 + 0.1703 + 0.2384 + 0.2225]
= 0.692
d)
When arrival distribution is Poisson(lambda), inter-arrival time distribution is
Exponential(lambda)
Thus, interarrival time distribution is Exponential(1.4)
Expected value of time between 5th and 6th ghost = 1/1.4 = 0.7143hr
e)
On average time taken for 10th ghost to arrive = 10*(1/1.4) = 7.143hr
Thus on an average, 10th ghost will arrive at 3:09pm
f)
Expected value of time between ghost appearances = 1/1.4 = 0.7143hr (see part d)
g)
Probability that 3rd ghost appears after 9:45 am
= Probablity that number of ghosts appearing till 9:45 am < 3
Duration from 8am to 9:45 am = 1.75hr
Mean number of ghosts = 1.4*1.75 = 2.45
Thus X is Poisson(2.45)
P(X < 3)
= P(X = 0) + P(X = 1) + P(X = 2)
= (2.45^0 * e^{-2.45} / 0!) + (2.45^1 * e^{-2.45} / 1!) + (2.45^2 * e^{-2.45} / 2!)
= 0.0863 + 0.2114 + 0.2590
= 0.5567
h)
As inter-arrival time is Exponential(1.4), inter-arrival time is memoryless
Thus, time for 7th ghost will be calculated from 1:00 pm and not from 12:35pm due to
memorylessness property.
Time duration from 1:00pm to 1:15pm = 0.25hrs
Let Y be random variable denoting inter-arrival time
probability that the 7th ghost will appear before 1:15 p.m.
= P(Y < 0.25)
= 1 - e^{-1.4 * 0.25)
= 0.2953
i)
Expected value of time for 7th ghost to appear after 1:00pm = 1/1.4 = 0.7143hr = 42.86 min
Time from 12:35pm to 1:00pm = 25min
Thus expected interarrival time = 25 + 42.86 = 67.86 min = 1hr 7.86min
j)
Expected duration after 1:00pm for 9th ghost to appear = 3 * (1/1.4) = 2.143hr = 2hr 8.58min
Thus expected time at which 9th ghost appears = 3:09pm
k)
As the inter-arrival time follows memorylessness property, the future probabilities do not change
with the observed value at past.
Thus, 4 ghosts appearing between 7:00pm to 10:00pm does not affect ghost appearance from
11:00pm to 11:30pm
Duration from 11:00pm to 11:30pm = 0.5hrs
mean number of ghosts = 1.4*0.5 = 0.7
Thus it follows Poisson(0.7) distribution
P(X > 2)
= 1 - P(X <= 2)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= 1 - [(0.7^0 * e^{-0.7} / 0!) + (0.7^1 * e^{-0.7} / 1!) + (0.7^2 * e^{-0.7} / 2!)]
= 1 - [0.4966 + 0.3476 + 0.1217]
= 0.0341
l)
Probability that 8 ghosts appear between 6pm and midnight
= Probability that number of ghosts appear from 6pm to 7pm and 10pm to midnight = 4 as it is
known that 4 ghosts appeared from 7pm to 10pm
Let A denote number of ghost from 6pm to 7pm
B denote number of ghosts .
2.a. Wired Media Type and ExplinationTwisted-Pair CableTwiste.pdf
2.
a. Wired Media Type and Explination:
Twisted-Pair Cable
Twisted-pair cable is a type of cabling that is used for telephone communications and most
modern Ethernet networks. A pair of wires forms a circuit that can transmit data. The pairs are
twisted to provide protection against crosstalk, the noise generated by adjacent pairs. When
electrical current flows through a wire, it creates a small, circular magnetic field around the wire.
When two wires in an electrical circuit are placed close together, their magnetic fields are the
exact opposite of each other. Thus, the two magnetic fields cancel each other out. They also
cancel out any outside magnetic fields. Twisting the wires can enhance this cancellation effect.
Using cancellation together with twisting the wires, cable designers can effectively provide self-
shielding for wire pairs within the network media.
Two basic types of twisted-pair cable exist: unshielded twisted pair (UTP) and shielded twisted
pair (STP). The following sections discuss UTP and STP cable in more detail.
UTP Cable
UTP cable is a medium that is composed of pairs of wires . UTP cable is used in a variety of
networks. Each of the eight individual copper wires in UTP cable \\is covered by an insulating
material. In addition, the wires in each pair are twisted around each other.
UTP cable relies solely on the cancellation effect produced by the twisted wire pairs to limit
signal degradation caused by electromagnetic interference (EMI) and radio frequency
interference (RFI). To further reduce crosstalk between the pairs in UTP cable, the number of
twists in the wire pairs varies. UTP cable must follow precise specifications governing how
many twists or braids are permitted per meter (3.28 feet) of cable.
UTP cable often is installed using a Registered Jack 45 (RJ-45) connector The RJ-45 is an eight-
wire connector used commonly to connect computers onto a local-area network (LAN),
especially Ethernets.
When used as a networking medium, UTP cable has four pairs of either 22- or 24-gauge copper
wire. UTP used as a networking medium has an impedance of 100 ohms; this differentiates it
from other types of twisted-pair wiring such as that used for telephone wiring, which has
impedance of 600 ohms.
UTP cable offers many advantages. Because UTP has an external diameter of approximately
0.43 cm (0.17 inches), its small size can be advantageous during installation. Because it has such
a small external diameter, UTP does not fill up wiring ducts as rapidly as other types of cable.
This can be an extremely important factor to consider, particularly when installing a network in
an older building. UTP cable is easy to install and is less expensive than other types of
networking media. In fact, UTP costs less per meter than any other type of LAN cabling. And
because UTP can be used with most of the major networking architectures, it continues to grow
in popularity.
Disadvantages also are involved in using twisted-pair cabling, howe.
A person may not choose to participate in the labour force due to La.pdf
A person may not choose to participate in the labour force due to Laziness, lack of ambition,
emotional or chemical imbalance, independent wealth, personal unavailability, or other priorities.
However, race, location, education, and marital status are irrelevant at times. They do affect the
type of job, or their availability, but not the individual\'s desire to have one. Choosing to stay
home and raise children is an example of different priorities, and a spouse\'s income, if they are
wealthy enough, may negate having to even make a choice.
(a) with low education, a job might not pay much more than unemployment benefit.
(b) with children, there is strong incentive to stay at home and take care of them.
(c) with one spouse earning a lot, another spouse can stay at home.
(d) if person belongs to a race with bad reputation, they are less likely to get hired. But at times
women are viewed as better workers than men from same race.
(e) in rural areas, there are fewer jobs, and the commute to them is longer and therefore they
prefer to do some small work nearby or choose not to work at all.
Solution
A person may not choose to participate in the labour force due to Laziness, lack of ambition,
emotional or chemical imbalance, independent wealth, personal unavailability, or other priorities.
However, race, location, education, and marital status are irrelevant at times. They do affect the
type of job, or their availability, but not the individual\'s desire to have one. Choosing to stay
home and raise children is an example of different priorities, and a spouse\'s income, if they are
wealthy enough, may negate having to even make a choice.
(a) with low education, a job might not pay much more than unemployment benefit.
(b) with children, there is strong incentive to stay at home and take care of them.
(c) with one spouse earning a lot, another spouse can stay at home.
(d) if person belongs to a race with bad reputation, they are less likely to get hired. But at times
women are viewed as better workers than men from same race.
(e) in rural areas, there are fewer jobs, and the commute to them is longer and therefore they
prefer to do some small work nearby or choose not to work at all..
Well.. 1) Ionic bonds are almost always metal to .pdf
Well.. 1) Ionic bonds are almost always metal to non-metal. (opposing charges
attract). 2) Covalent, electron-sharing. \"co\"-valent. Valent referring to the valence electrons of
the element, that is, the electrons in outer most shell. Remember each element wants to reach a
more stable state. To achieve this it will share electrons with another element through covalent
bonds. 3) a covalent bond is said to be polar if one of the elements sharing the electrons has a
greater electronegativity. (i.e. water contains polar covalent bonds between the oxygen and the
hydrogens because oxygen is far more electronegative than hydrogen, therefore it \"hogs\" the
electrons. 4) non-polar covalent bonds have an equal sharing of the electrons, equal
electronegativity. Hope this helps!
Solution
Well.. 1) Ionic bonds are almost always metal to non-metal. (opposing charges
attract). 2) Covalent, electron-sharing. \"co\"-valent. Valent referring to the valence electrons of
the element, that is, the electrons in outer most shell. Remember each element wants to reach a
more stable state. To achieve this it will share electrons with another element through covalent
bonds. 3) a covalent bond is said to be polar if one of the elements sharing the electrons has a
greater electronegativity. (i.e. water contains polar covalent bonds between the oxygen and the
hydrogens because oxygen is far more electronegative than hydrogen, therefore it \"hogs\" the
electrons. 4) non-polar covalent bonds have an equal sharing of the electrons, equal
electronegativity. Hope this helps!.
1. The answer is d) Environmental EffectsEnvironmental effects ca.pdf
1. The answer is: d) Environmental Effects
Environmental effects can cause a single genotype to produce a range of potential Phenotypes.
2. The answer is: c) Differ in both mean and variance
The Figure represents two distributions that differ in both mean and variance.
Solution
1. The answer is: d) Environmental Effects
Environmental effects can cause a single genotype to produce a range of potential Phenotypes.
2. The answer is: c) Differ in both mean and variance
The Figure represents two distributions that differ in both mean and variance..
clear clc close all Use polyfit to solve for the phase l.pdf
clear
clc
close all
%Use polyfit to solve for the phase lines
point11=[0,15];
point12=[0,300];
line1=polyfit(point11,point12,1);
m1=line1(1);
b1=line1(2);
xpf1=linspace(min(point11),max(point11),100);
ypf1=m1*xpf1+b1;
point21=[0,15];
point22=[700,300];
line2=polyfit(point21,point22,1);
m2=line2(1);
b2=line2(2);
xpf2=linspace(min(point21),max(point21),100);
ypf2=m2*xpf2+b2;
point31=[0,50];
point32=[700,300];
line3=polyfit(point31,point32,1);
m3=line3(1);
b3=line3(2);
xpf3=linspace(min(point31),max(point31),100);
ypf3=m3*xpf3+b3;
point41=[50,100];
point42=[300,800];
line4=polyfit(point41,point42,1);
m4=line4(1);
b4=line4(2);
xpf4=linspace(min(point41),max(point41),100);
ypf4=m4*xpf4+b4;
point51=[100,85];
point52=[800,300];
line5=polyfit(point51,point52,1);
m5=line5(1);
b5=line5(2);
xpf5=linspace(min(point51),max(point51),100);
ypf5=m5*xpf5+b5;
point61=[85,100];
point62=[300,0];
line6=polyfit(point61,point62,1);
m6=line6(1);
b6=line6(2);
xpf6=linspace(min(point61),max(point61),100);
ypf6=m6*xpf6+b6;
point71=[15,85];
point72=[300,300];
%Allow the user to set the percent of B
b=input(\'Enter the mass percent of B: \');
%Return an error if the percent of B is not a valid number
if b<0 || b>100
error(\'Please enter a value between 0 and 100.\')
end
%Allow the user to set the temperature
t=input(\'Enter the temperature [deg C]: \');
%Solve for percent composition of A
a=100-b;
%Determine phase based on temperature and percent composition of B using
%the phase lines
if t<300
if b<15 && t>m1*b+b1
phase=\'alpha\';
elseif b<15 && t=15 && b<=85
phase=\'alpha + beta\';
elseif b>85 && t>m6*b+b6
phase=\'beta\';
else phase=\'alpha + beta\';
end
elseif b<15
if t=15 && b<=50
if t50 && b<=85
if t=15 && b<=85 && t==300
fprintf(\'The provided conditions fall on the eutectic line.\ \')
end
if b==50 && t==300
fprintf(\'The provided conditions fall on the eutectic point.\ \')
end
%Output a formatted statement to display the phase
fprintf(\'For the composition of %0.2f%% A, %0.2f%% B and a temperature of %0.0f degrees
Celsius, the phase is %s.\ \',a,b,t,phase)
%Create a plot of the phase lines and the user\'s set point
figure(\'color\',\'white\')
%Set axes
axis([0 100 0 1000])
%Plot the phase lines, eutectic line, eutectic point, and the user\'s point
plot(xpf1,ypf1,\'-b\')
hold on
plot(xpf2,ypf2,\'-b\')
plot(xpf3,ypf3,\'-b\')
plot(xpf4,ypf4,\'-b\')
plot(xpf5,ypf5,\'-b\')
plot(xpf6,ypf6,\'-b\')
plot(point71,point72,\'-k\')
plot(b,t,\'ro\',\'MarkerFaceColor\',\'r\')
%Assign a title and axes labels
title(\'Phase Diagram of Elements A and B\')
xlabel(\'Percent Composition of B\')
ylabel(\'Temperature (T) [deg C]\')
%Set scaling for the axes
set(gca,\'XTick\',0:10:100,\'YTick\',0:100:1000)
%Place the phase names on the graph
text(40,200,\'alpha+beta\')
text(45,600,\'Liquid\')
text(90,300,\'beta\')
text(3,300,\'alpha\')
text(15,375,\'alpha + Liquid\')
text(62,375,\'beta + Liquid\')
Solution
clear
clc
close all
%Use polyfit to solve for the phase lines
point11=[0,15];
point.
The oxygen appears in both step reactions. But, i.pdf
The oxygen appears in both step reactions. But, it would not appear in the final
reaction. Therefore it is a reaction intermediate. It is sort of in the \"middle\" The final reaction
is 2O3 ====> 3 O2 (No O)
Solution
The oxygen appears in both step reactions. But, it would not appear in the final
reaction. Therefore it is a reaction intermediate. It is sort of in the \"middle\" The final reaction
is 2O3 ====> 3 O2 (No O).
The two contributions to the cohesive energy of t.pdf
The two contributions to the cohesive energy of the ionic crystal are : Vander waals
attraction between the adjacent ions Zero point oscillations of their equilibrium position. Zero-
point energy is the lowest possible energy that a quantum mechanical physical system may have;
it is the energy of its ground state. All quantum mechanical systems undergo fluctuations even in
their ground state and have an associated zero-point energy, a consequence of their wave-like
interaction.
Solution
The two contributions to the cohesive energy of the ionic crystal are : Vander waals
attraction between the adjacent ions Zero point oscillations of their equilibrium position. Zero-
point energy is the lowest possible energy that a quantum mechanical physical system may have;
it is the energy of its ground state. All quantum mechanical systems undergo fluctuations even in
their ground state and have an associated zero-point energy, a consequence of their wave-like
interaction..
If you are talking about an extraction design, th.pdf
If you are talking about an extraction design, the the water solubility has to be
greater in the aqueous base because of the greater solubility of the salts of the acids (with few
exceptions), again presuming a differential extraction aqueous to organic.
Solution
If you are talking about an extraction design, the the water solubility has to be
greater in the aqueous base because of the greater solubility of the salts of the acids (with few
exceptions), again presuming a differential extraction aqueous to organic..
Yes ,its true. Though both gibbons and rhesus monkeys belong to pr.pdf
Yes ,it\'s true. Though both gibbons and rhesus monkeys belong to primates, but gibbons are
under a group apes, and rhesus monkeys belongs to old world monkey group. Apes are near
common ancestor of homosapiens.
Solution
Yes ,it\'s true. Though both gibbons and rhesus monkeys belong to primates, but gibbons are
under a group apes, and rhesus monkeys belongs to old world monkey group. Apes are near
common ancestor of homosapiens..
More Related Content
Similar to PART A The element Si belongs to IVA group. Therefore, four electro.pdf
Class XII d and f- block elements (part 2)
This part contains ionisation enthalpies, oxidation states, metal oxides & oxocations, magnetic properties, coloured ions of d block elements, catalysts, interstitial compounds, alloy formation & some important conceptual questions with answers as hints. Also some reasoning questions are given to test the understanding of properties of d block elements.
Similar to PART A The element Si belongs to IVA group. Therefore, four electro.pdf (8)
4structureoftheatom-150805145649-lva1-app6892.pptx
4structureoftheatom-150805145649-lva1-app6892.pptx
More from aquazac
Answer Every investor expects dividend from his investments.Dividen.pdf
Answer: Every investor expects dividend from his investments.Dividend income helps investors
in dealing with their routine expenses. Investors put their money to earn some positive returns
and dividend is one of the way of getting returns for investors. So, It is a duty of management to
satisfy dividend expectation of investors.
If management do not pay dividend then it is known as retention of earnings. Retention earnings
has it\'s own pro and cons, which are described as follows:
Pros:
Retained earnings reduces organization\'s dependence on external borrowings. It can be
considered as an advantage when external borrowings are not available easily.
Retained earnings indicates good business opportunities, It means organization is growing
financially. It may result in better stock valuation of organization.
In some situation dividend is subject to taxes (either distribution tax or income tax) in the hands
of investors. So, in this case investors are not interested in dividend but they are more interested
in capital gain(which may be tax exempt). So, In this situation retained earnings fulfill investor\'s
objective.
Cons:
As explained earlier all the investors have some expectation of dividend, in case of higher
retention of earnings it may become difficult to fulfill investor\'s expectations.
Retention of earnings may affect investor\'s earning capacity when investors have available
opportunities, which gives higher return than organization earns by retaining amount. In this case
investor looses earning opportunities.
In case of dividend management\'s decision is significantly influenced by media and institutional
investors. Media is responsible for organization\'s image in society, such image significantly
affects organization\'s financial performance. This is why management always considers media\'s
point of view before making any dividend decision.
Institutional investors are usually invests their money and takes part in decision making process
of organization. Institutional investor\'s weight in decision making process is generally very
high, so it is obvious that they influence management\'s dividend decision also.
Solution
Answer: Every investor expects dividend from his investments.Dividend income helps investors
in dealing with their routine expenses. Investors put their money to earn some positive returns
and dividend is one of the way of getting returns for investors. So, It is a duty of management to
satisfy dividend expectation of investors.
If management do not pay dividend then it is known as retention of earnings. Retention earnings
has it\'s own pro and cons, which are described as follows:
Pros:
Retained earnings reduces organization\'s dependence on external borrowings. It can be
considered as an advantage when external borrowings are not available easily.
Retained earnings indicates good business opportunities, It means organization is growing
financially. It may result in better stock valuation of organization.
In some situ.
Ans. Gene is defined as the segment of DNA that gives a functional p.pdf
Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A
functional product can be either protein (or polypeptide) or RNA.
mRNA transcribes from a gene is further translated to produce polypeptides. However, all other
RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and
various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid
carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA
mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene
regulations.
SO, polypeptides are just one class of functional product from DNA. The genomic DNA also
produces various classes of RNAs with diverse functions that are crucial for cellular activates
and gene regulation.
Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides
can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all
genes (including those encoding polypeptides and RNAs).
Solution
Ans. Gene is defined as the segment of DNA that gives a functional product (protein, RNA). A
functional product can be either protein (or polypeptide) or RNA.
mRNA transcribes from a gene is further translated to produce polypeptides. However, all other
RNAs are also transcribed from their respective genes. It includes tRNA, rRNA, snRNA and
various regulatory RNAs (micro RNA, small interfering RNA), etc. tRNA acts as amino-acid
carrier during translation. rRNA is the structural and functional scaffold of ribosomes. snRNA
mediates splicing of immature mRNA. miRNA and siRNA have crucial roles in gene
regulations.
SO, polypeptides are just one class of functional product from DNA. The genomic DNA also
produces various classes of RNAs with diverse functions that are crucial for cellular activates
and gene regulation.
Therefore, the statement “Add DNA sequences in the genome that does not encode polypeptides
can be defined as junk DNA” is INCORRECT. Junk DNA is non-coding DNA that excludes all
genes (including those encoding polypeptides and RNAs)..
additional optmization techniques for underlying IP network must1.pdf
additional optmization techniques for underlying IP network must:
1.Have bandwidth and performance to handle converged services.
2.Seprate VLAN for Voice traffic.
3.Meet the demand of high availability of voice services by providing resiliency to mitigate the
effect of network outages.
4.Should be modular, hierarchical & consistent to promote consistency and manageability.
5.configure QoS on core switch.
Solution
additional optmization techniques for underlying IP network must:
1.Have bandwidth and performance to handle converged services.
2.Seprate VLAN for Voice traffic.
3.Meet the demand of high availability of voice services by providing resiliency to mitigate the
effect of network outages.
4.Should be modular, hierarchical & consistent to promote consistency and manageability.
5.configure QoS on core switch..
According to the given equation, aqueous carbon dioxide reacts with .pdf
According to the given equation, aqueous carbon dioxide reacts with water to form carbonic
acid, Carbonic acid is not stable in normal body temperatures, therefore, it ionizes into
bicarbonate ion (HCO3-) and a hydrogen ion (H+) is liberated. The liberation of hydrogen ions
decreases the pH.
During breathing, oxygen is inhaled which is carried by the blood to the different tissues of the
body. Oxygen facilitates the various metabolic reactions occurring inside our cells and helps in
respiration, or the generation of energy. Carbon dioxide, formed as a by-product of these energy-
generating metabolic reactions, diffuses out of the tissues and into the blood stream, which
carries CO2 to the lungs, where it is exhaled out.
a) In case of hypoventilation :
The exchange of gases that occurs in the lungs while breathing is impaired. Therefore, the
requisite amount of carbon dioxide is not being exhaled out. This leads to a build-up of carbon-
dioxide in the blood plasma. As a result, the forward reaction is facilitated and more H+ ions are
generated. The higher concentration of H+ ions decreases the pH (acidic pH). This can lead to
acidosis.
The build-up of carbon dioxide in the blood facilitates the forward reaction and therefore the
equilibrium shifts to the right.
The hydronium ion or H3O+ is the product of a hydrogen ion (H+) and water. Since a hydrogen
ion is too small to exist freely in water, it therefore combines with a water molecule to form a
hydronium ion which is more stable. In case of hypoventilation, the concentration of H+ ions
increase. Therefore, the concentration of hydronium ions increase during hypoventilation.
The stress concentration or partial pressure of CO2 during hypoventilation is significantly higher
than 40mmHg (around 60-65 mmHg).
b) In case of hyperventilation :
The breathing rate is more than normal. Therefore, as a result of heavier and faster breathing, the
CO2 concentration in the blood rapidly decreases, that is, CO2 easily leaves the blood plasma.
Decrease in the CO2 concentration slows down the rate of the forward reaction, reducing the
levels at which bicarbonate ion and H+ ion concentration were being generated. The bicarbonate
and hydrogen ions recombine to yield CO2 and water. This lowering in hydrogen ion
concentration consequently lowers the pH (alkaline pH). This can lead to alkalosis.
As a result of the lowering of CO2 , the reverse reaction takes place. The carbonic acid changes
back to CO2 and H2O, shifting the equilibrium to the left.
The concentration of H3O+ decreases during hyperventilation, since the H+ ion concentration is
lowered.
The stress levels or partial pressure of CO2 during hyperventilation is significantly lower than 40
mmHg (around 25-30 mmHg).
c) If sodium carbonate or Na2CO3 is added instead of sodium bicarbonate, then sodium
carbonate would react with water and it would hydrolyse into Na+ and CO32- ions. The CO32-
combine with the hydrogen (H) present in water to form H2CO3 (carbon.
a) mean = 1.43Thus distribution is Poisson(4.2)P(X = 4) = 4.2^4.pdf
a) mean = 1.4*3
Thus distribution is Poisson(4.2)
P(X = 4) = 4.2^4 * e^{-4.2} / 4! = 0.1944
b)
average number of ghosts in 90min (1.5hrs) interval = 1.4*1.5 = 2.1
c)
Probability that 4th ghost will appear before 10 am
= Probability that number of ghosts appearing from 8 to 10 am >= 4
As it is a 2 hr period, the distribution of number of ghosts is Poisson(1.4*2) i.e Poisson(2.8)
P(X >= 4)
= 1 - P(X < 4)
= 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]
= 1 - [(2.8^0 * e^{-2.8} / 0!) + (2.8^1 * e^{-2.8} / 1!) + (2.8^2 * e^{-2.8} / 2!) + (2.8^3 * e^{-
2.8} / 3!)]
= 1 - [0.0608 + 0.1703 + 0.2384 + 0.2225]
= 0.692
d)
When arrival distribution is Poisson(lambda), inter-arrival time distribution is
Exponential(lambda)
Thus, interarrival time distribution is Exponential(1.4)
Expected value of time between 5th and 6th ghost = 1/1.4 = 0.7143hr
e)
On average time taken for 10th ghost to arrive = 10*(1/1.4) = 7.143hr
Thus on an average, 10th ghost will arrive at 3:09pm
f)
Expected value of time between ghost appearances = 1/1.4 = 0.7143hr (see part d)
g)
Probability that 3rd ghost appears after 9:45 am
= Probablity that number of ghosts appearing till 9:45 am < 3
Duration from 8am to 9:45 am = 1.75hr
Mean number of ghosts = 1.4*1.75 = 2.45
Thus X is Poisson(2.45)
P(X < 3)
= P(X = 0) + P(X = 1) + P(X = 2)
= (2.45^0 * e^{-2.45} / 0!) + (2.45^1 * e^{-2.45} / 1!) + (2.45^2 * e^{-2.45} / 2!)
= 0.0863 + 0.2114 + 0.2590
= 0.5567
h)
As inter-arrival time is Exponential(1.4), inter-arrival time is memoryless
Thus, time for 7th ghost will be calculated from 1:00 pm and not from 12:35pm due to
memorylessness property.
Time duration from 1:00pm to 1:15pm = 0.25hrs
Let Y be random variable denoting inter-arrival time
probability that the 7th ghost will appear before 1:15 p.m.
= P(Y < 0.25)
= 1 - e^{-1.4 * 0.25)
= 0.2953
i)
Expected value of time for 7th ghost to appear after 1:00pm = 1/1.4 = 0.7143hr = 42.86 min
Time from 12:35pm to 1:00pm = 25min
Thus expected interarrival time = 25 + 42.86 = 67.86 min = 1hr 7.86min
j)
Expected duration after 1:00pm for 9th ghost to appear = 3 * (1/1.4) = 2.143hr = 2hr 8.58min
Thus expected time at which 9th ghost appears = 3:09pm
k)
As the inter-arrival time follows memorylessness property, the future probabilities do not change
with the observed value at past.
Thus, 4 ghosts appearing between 7:00pm to 10:00pm does not affect ghost appearance from
11:00pm to 11:30pm
Duration from 11:00pm to 11:30pm = 0.5hrs
mean number of ghosts = 1.4*0.5 = 0.7
Thus it follows Poisson(0.7) distribution
P(X > 2)
= 1 - P(X <= 2)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= 1 - [(0.7^0 * e^{-0.7} / 0!) + (0.7^1 * e^{-0.7} / 1!) + (0.7^2 * e^{-0.7} / 2!)]
= 1 - [0.4966 + 0.3476 + 0.1217]
= 0.0341
l)
Probability that 8 ghosts appear between 6pm and midnight
= Probability that number of ghosts appear from 6pm to 7pm and 10pm to midnight = 4 as it is
known that 4 ghosts appeared from 7pm to 10pm
Let A denote number of ghost from 6pm to 7pm
B denote number of ghosts .
2.a. Wired Media Type and ExplinationTwisted-Pair CableTwiste.pdf
2.
a. Wired Media Type and Explination:
Twisted-Pair Cable
Twisted-pair cable is a type of cabling that is used for telephone communications and most
modern Ethernet networks. A pair of wires forms a circuit that can transmit data. The pairs are
twisted to provide protection against crosstalk, the noise generated by adjacent pairs. When
electrical current flows through a wire, it creates a small, circular magnetic field around the wire.
When two wires in an electrical circuit are placed close together, their magnetic fields are the
exact opposite of each other. Thus, the two magnetic fields cancel each other out. They also
cancel out any outside magnetic fields. Twisting the wires can enhance this cancellation effect.
Using cancellation together with twisting the wires, cable designers can effectively provide self-
shielding for wire pairs within the network media.
Two basic types of twisted-pair cable exist: unshielded twisted pair (UTP) and shielded twisted
pair (STP). The following sections discuss UTP and STP cable in more detail.
UTP Cable
UTP cable is a medium that is composed of pairs of wires . UTP cable is used in a variety of
networks. Each of the eight individual copper wires in UTP cable \\is covered by an insulating
material. In addition, the wires in each pair are twisted around each other.
UTP cable relies solely on the cancellation effect produced by the twisted wire pairs to limit
signal degradation caused by electromagnetic interference (EMI) and radio frequency
interference (RFI). To further reduce crosstalk between the pairs in UTP cable, the number of
twists in the wire pairs varies. UTP cable must follow precise specifications governing how
many twists or braids are permitted per meter (3.28 feet) of cable.
UTP cable often is installed using a Registered Jack 45 (RJ-45) connector The RJ-45 is an eight-
wire connector used commonly to connect computers onto a local-area network (LAN),
especially Ethernets.
When used as a networking medium, UTP cable has four pairs of either 22- or 24-gauge copper
wire. UTP used as a networking medium has an impedance of 100 ohms; this differentiates it
from other types of twisted-pair wiring such as that used for telephone wiring, which has
impedance of 600 ohms.
UTP cable offers many advantages. Because UTP has an external diameter of approximately
0.43 cm (0.17 inches), its small size can be advantageous during installation. Because it has such
a small external diameter, UTP does not fill up wiring ducts as rapidly as other types of cable.
This can be an extremely important factor to consider, particularly when installing a network in
an older building. UTP cable is easy to install and is less expensive than other types of
networking media. In fact, UTP costs less per meter than any other type of LAN cabling. And
because UTP can be used with most of the major networking architectures, it continues to grow
in popularity.
Disadvantages also are involved in using twisted-pair cabling, howe.
A person may not choose to participate in the labour force due to La.pdf
A person may not choose to participate in the labour force due to Laziness, lack of ambition,
emotional or chemical imbalance, independent wealth, personal unavailability, or other priorities.
However, race, location, education, and marital status are irrelevant at times. They do affect the
type of job, or their availability, but not the individual\'s desire to have one. Choosing to stay
home and raise children is an example of different priorities, and a spouse\'s income, if they are
wealthy enough, may negate having to even make a choice.
(a) with low education, a job might not pay much more than unemployment benefit.
(b) with children, there is strong incentive to stay at home and take care of them.
(c) with one spouse earning a lot, another spouse can stay at home.
(d) if person belongs to a race with bad reputation, they are less likely to get hired. But at times
women are viewed as better workers than men from same race.
(e) in rural areas, there are fewer jobs, and the commute to them is longer and therefore they
prefer to do some small work nearby or choose not to work at all.
Solution
A person may not choose to participate in the labour force due to Laziness, lack of ambition,
emotional or chemical imbalance, independent wealth, personal unavailability, or other priorities.
However, race, location, education, and marital status are irrelevant at times. They do affect the
type of job, or their availability, but not the individual\'s desire to have one. Choosing to stay
home and raise children is an example of different priorities, and a spouse\'s income, if they are
wealthy enough, may negate having to even make a choice.
(a) with low education, a job might not pay much more than unemployment benefit.
(b) with children, there is strong incentive to stay at home and take care of them.
(c) with one spouse earning a lot, another spouse can stay at home.
(d) if person belongs to a race with bad reputation, they are less likely to get hired. But at times
women are viewed as better workers than men from same race.
(e) in rural areas, there are fewer jobs, and the commute to them is longer and therefore they
prefer to do some small work nearby or choose not to work at all..
Well.. 1) Ionic bonds are almost always metal to .pdf
Well.. 1) Ionic bonds are almost always metal to non-metal. (opposing charges
attract). 2) Covalent, electron-sharing. \"co\"-valent. Valent referring to the valence electrons of
the element, that is, the electrons in outer most shell. Remember each element wants to reach a
more stable state. To achieve this it will share electrons with another element through covalent
bonds. 3) a covalent bond is said to be polar if one of the elements sharing the electrons has a
greater electronegativity. (i.e. water contains polar covalent bonds between the oxygen and the
hydrogens because oxygen is far more electronegative than hydrogen, therefore it \"hogs\" the
electrons. 4) non-polar covalent bonds have an equal sharing of the electrons, equal
electronegativity. Hope this helps!
Solution
Well.. 1) Ionic bonds are almost always metal to non-metal. (opposing charges
attract). 2) Covalent, electron-sharing. \"co\"-valent. Valent referring to the valence electrons of
the element, that is, the electrons in outer most shell. Remember each element wants to reach a
more stable state. To achieve this it will share electrons with another element through covalent
bonds. 3) a covalent bond is said to be polar if one of the elements sharing the electrons has a
greater electronegativity. (i.e. water contains polar covalent bonds between the oxygen and the
hydrogens because oxygen is far more electronegative than hydrogen, therefore it \"hogs\" the
electrons. 4) non-polar covalent bonds have an equal sharing of the electrons, equal
electronegativity. Hope this helps!.
1. The answer is d) Environmental EffectsEnvironmental effects ca.pdf
1. The answer is: d) Environmental Effects
Environmental effects can cause a single genotype to produce a range of potential Phenotypes.
2. The answer is: c) Differ in both mean and variance
The Figure represents two distributions that differ in both mean and variance.
Solution
1. The answer is: d) Environmental Effects
Environmental effects can cause a single genotype to produce a range of potential Phenotypes.
2. The answer is: c) Differ in both mean and variance
The Figure represents two distributions that differ in both mean and variance..
clear clc close all Use polyfit to solve for the phase l.pdf
clear
clc
close all
%Use polyfit to solve for the phase lines
point11=[0,15];
point12=[0,300];
line1=polyfit(point11,point12,1);
m1=line1(1);
b1=line1(2);
xpf1=linspace(min(point11),max(point11),100);
ypf1=m1*xpf1+b1;
point21=[0,15];
point22=[700,300];
line2=polyfit(point21,point22,1);
m2=line2(1);
b2=line2(2);
xpf2=linspace(min(point21),max(point21),100);
ypf2=m2*xpf2+b2;
point31=[0,50];
point32=[700,300];
line3=polyfit(point31,point32,1);
m3=line3(1);
b3=line3(2);
xpf3=linspace(min(point31),max(point31),100);
ypf3=m3*xpf3+b3;
point41=[50,100];
point42=[300,800];
line4=polyfit(point41,point42,1);
m4=line4(1);
b4=line4(2);
xpf4=linspace(min(point41),max(point41),100);
ypf4=m4*xpf4+b4;
point51=[100,85];
point52=[800,300];
line5=polyfit(point51,point52,1);
m5=line5(1);
b5=line5(2);
xpf5=linspace(min(point51),max(point51),100);
ypf5=m5*xpf5+b5;
point61=[85,100];
point62=[300,0];
line6=polyfit(point61,point62,1);
m6=line6(1);
b6=line6(2);
xpf6=linspace(min(point61),max(point61),100);
ypf6=m6*xpf6+b6;
point71=[15,85];
point72=[300,300];
%Allow the user to set the percent of B
b=input(\'Enter the mass percent of B: \');
%Return an error if the percent of B is not a valid number
if b<0 || b>100
error(\'Please enter a value between 0 and 100.\')
end
%Allow the user to set the temperature
t=input(\'Enter the temperature [deg C]: \');
%Solve for percent composition of A
a=100-b;
%Determine phase based on temperature and percent composition of B using
%the phase lines
if t<300
if b<15 && t>m1*b+b1
phase=\'alpha\';
elseif b<15 && t=15 && b<=85
phase=\'alpha + beta\';
elseif b>85 && t>m6*b+b6
phase=\'beta\';
else phase=\'alpha + beta\';
end
elseif b<15
if t=15 && b<=50
if t50 && b<=85
if t=15 && b<=85 && t==300
fprintf(\'The provided conditions fall on the eutectic line.\ \')
end
if b==50 && t==300
fprintf(\'The provided conditions fall on the eutectic point.\ \')
end
%Output a formatted statement to display the phase
fprintf(\'For the composition of %0.2f%% A, %0.2f%% B and a temperature of %0.0f degrees
Celsius, the phase is %s.\ \',a,b,t,phase)
%Create a plot of the phase lines and the user\'s set point
figure(\'color\',\'white\')
%Set axes
axis([0 100 0 1000])
%Plot the phase lines, eutectic line, eutectic point, and the user\'s point
plot(xpf1,ypf1,\'-b\')
hold on
plot(xpf2,ypf2,\'-b\')
plot(xpf3,ypf3,\'-b\')
plot(xpf4,ypf4,\'-b\')
plot(xpf5,ypf5,\'-b\')
plot(xpf6,ypf6,\'-b\')
plot(point71,point72,\'-k\')
plot(b,t,\'ro\',\'MarkerFaceColor\',\'r\')
%Assign a title and axes labels
title(\'Phase Diagram of Elements A and B\')
xlabel(\'Percent Composition of B\')
ylabel(\'Temperature (T) [deg C]\')
%Set scaling for the axes
set(gca,\'XTick\',0:10:100,\'YTick\',0:100:1000)
%Place the phase names on the graph
text(40,200,\'alpha+beta\')
text(45,600,\'Liquid\')
text(90,300,\'beta\')
text(3,300,\'alpha\')
text(15,375,\'alpha + Liquid\')
text(62,375,\'beta + Liquid\')
Solution
clear
clc
close all
%Use polyfit to solve for the phase lines
point11=[0,15];
point.
The oxygen appears in both step reactions. But, i.pdf
The oxygen appears in both step reactions. But, it would not appear in the final
reaction. Therefore it is a reaction intermediate. It is sort of in the \"middle\" The final reaction
is 2O3 ====> 3 O2 (No O)
Solution
The oxygen appears in both step reactions. But, it would not appear in the final
reaction. Therefore it is a reaction intermediate. It is sort of in the \"middle\" The final reaction
is 2O3 ====> 3 O2 (No O).
The two contributions to the cohesive energy of t.pdf
The two contributions to the cohesive energy of the ionic crystal are : Vander waals
attraction between the adjacent ions Zero point oscillations of their equilibrium position. Zero-
point energy is the lowest possible energy that a quantum mechanical physical system may have;
it is the energy of its ground state. All quantum mechanical systems undergo fluctuations even in
their ground state and have an associated zero-point energy, a consequence of their wave-like
interaction.
Solution
The two contributions to the cohesive energy of the ionic crystal are : Vander waals
attraction between the adjacent ions Zero point oscillations of their equilibrium position. Zero-
point energy is the lowest possible energy that a quantum mechanical physical system may have;
it is the energy of its ground state. All quantum mechanical systems undergo fluctuations even in
their ground state and have an associated zero-point energy, a consequence of their wave-like
interaction..
If you are talking about an extraction design, th.pdf
If you are talking about an extraction design, the the water solubility has to be
greater in the aqueous base because of the greater solubility of the salts of the acids (with few
exceptions), again presuming a differential extraction aqueous to organic.
Solution
If you are talking about an extraction design, the the water solubility has to be
greater in the aqueous base because of the greater solubility of the salts of the acids (with few
exceptions), again presuming a differential extraction aqueous to organic..
Yes ,its true. Though both gibbons and rhesus monkeys belong to pr.pdf
Yes ,it\'s true. Though both gibbons and rhesus monkeys belong to primates, but gibbons are
under a group apes, and rhesus monkeys belongs to old world monkey group. Apes are near
common ancestor of homosapiens.
Solution
Yes ,it\'s true. Though both gibbons and rhesus monkeys belong to primates, but gibbons are
under a group apes, and rhesus monkeys belongs to old world monkey group. Apes are near
common ancestor of homosapiens..
When something boils, it changes states of matter. It would go from .pdf
When something boils, it changes states of matter. It would go from a liquid to a soild when it
boils, so if more heat is added, it will just change the state of matter faster.
Solution
When something boils, it changes states of matter. It would go from a liquid to a soild when it
boils, so if more heat is added, it will just change the state of matter faster..
We need to discuss why there is an importance of adding residents to.pdf
We need to discuss why there is an importance of adding residents to the budget planning, we
can go through below explaintion and can derived various conclusions from the same.
In explaining why they did or did not seek input from citizens, respondents in- dicated that the
process was guided by the desires of the leadership. Of those who said that they did seek citizen
input, the desire (as opposed to the obligation through policy) of the staff or the govern- ing
board or staff to seek public opinion seemed to provide some motivation. Almost three-quarters
of responding cities cited staff desire, and almost two-thirds, governing board desire.
The pattern for counties was similar. The most often cited reason for extra effort to involve the
public was board interest. Almost half of the counties also cited staff interest.
Although citizen involvement is a hot topic in management literature these days, local
governments in North Caro- lina are generally not making extensive fforts to involve residents in
budgeting. Barely half of the cities responding to the survey and less than a third of the coun- ties
go beyond the single, mandated hearing to get citizen input In general, cities tend to seek input
more than counties do. This result might be explained by the different types of services offered
by the two forms of gov- ernment. Counties provide funding for, among other things, social
services, pub- lic health services, mental health ser- vices, and schools, and the boards over-
seeing these services may seek citizen input when they are preparing their own budget request
for the county.
Although there is informal interest in hearing from citizens, boards clearly want to remain
flexible in when and how they obtain input. Only 13 percent of the city respondents had a formal
pol- icy or requirement for citizen involve- ment beyond the single, state-mandated hearing.
Medium to large cities were more likely to have such requirements than small cities. Only one
county, a larger one, reported having such a policy or requirement.
Although citizen involvement is a hot topic in management literature these days, local
governments in North Caro- lina are generally not making extensive
efforts to involve residents in budgeting. Barely half of the cities responding to the survey and
less than a third of the coun- ties go beyond the single, mandated hearing to get citizen input
Of those that did not seek extra citizen involvement, the most common reason cited, on both the
city and the county level, was lack of governing board inter- est. The second most common
reason was poor response in the past, when at- tempts to get citizens involved did not seem very
effective. At the county level, lack of board interest was a far more important factor than
anything else, including poor past response: 71 percent of respondents cited lack of board
interest, whereas only 39 percent cited poor past response Although staff interest seemed to be a
big motivator for involving ci.
What is the largest decimal integer that can be represented with the.pdf
What is the largest decimal integer that can be represented with the following number of bits?
1. 3 bits
2. 8 bits
3. 32 bits
4. 64 bits
How do you find out the largest decimal integer?
Solution
In general, with k bits we can represent 2k values. Their range will depend on the system:
Unsigned: 0 to 2k-1
1. 3 bits
=> 2^3 - 1 => 7 => 111
2. 8 bits
=> 2^8 - 1 => 255
3. 32 bits
=> 2^32 - 1
4. 64 bits
=> 2^64 - 1.
Throwing.javaimport java.util.InputMismatchException; import jav.pdf
Throwing.java
import java.util.InputMismatchException;
import java.util.Scanner;
public class Throwing
{
public static void main(String[] args)
{
try{
int n = getInt();
System.out.println(n);
}
catch(IllegalArgumentException e){
System.out.println(e.getMessage());
}
}
public static int getInt()
{
// your code goes here
int n;
try{
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter an integer: \");
n = scan.nextInt();
}
catch(InputMismatchException e){
throw new IllegalArgumentException(\"Invalid Input. Please enter an integer.\");
}
return n;
}
}
Output:
Enter an integer:
s
Invalid Input. Please enter an integer.
Throwing.java
import java.io.IOException;
import java.util.InputMismatchException;
import java.util.Scanner;
public class Throwing
{
public static void main(String[] args)
{
try{
int n = getInt();
System.out.println(n);
}
catch(IOException e){
System.out.println(e.getMessage());
}
}
public static int getInt() throws IOException
{
// your code goes here
int n;
try{
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter an integer: \");
n = scan.nextInt();
}
catch(InputMismatchException e){
throw new IOException(\"Invalid Input. Please enter an integer.\");
}
return n;
}
}
Output:
Enter an integer:
d
Invalid Input. Please enter an integer.
Solution
Throwing.java
import java.util.InputMismatchException;
import java.util.Scanner;
public class Throwing
{
public static void main(String[] args)
{
try{
int n = getInt();
System.out.println(n);
}
catch(IllegalArgumentException e){
System.out.println(e.getMessage());
}
}
public static int getInt()
{
// your code goes here
int n;
try{
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter an integer: \");
n = scan.nextInt();
}
catch(InputMismatchException e){
throw new IllegalArgumentException(\"Invalid Input. Please enter an integer.\");
}
return n;
}
}
Output:
Enter an integer:
s
Invalid Input. Please enter an integer.
Throwing.java
import java.io.IOException;
import java.util.InputMismatchException;
import java.util.Scanner;
public class Throwing
{
public static void main(String[] args)
{
try{
int n = getInt();
System.out.println(n);
}
catch(IOException e){
System.out.println(e.getMessage());
}
}
public static int getInt() throws IOException
{
// your code goes here
int n;
try{
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter an integer: \");
n = scan.nextInt();
}
catch(InputMismatchException e){
throw new IOException(\"Invalid Input. Please enter an integer.\");
}
return n;
}
}
Output:
Enter an integer:
d
Invalid Input. Please enter an integer..
This electron transport is accompanied by the protons transfer into .pdf
This electron transport is accompanied by the protons transfer into intermembrane space.
Protons release potential which drives the phosphorylation of ADP to ATP.
Protons return to the matrix through a proton specific channel
Solution
This electron transport is accompanied by the protons transfer into intermembrane space.
Protons release potential which drives the phosphorylation of ADP to ATP.
Protons return to the matrix through a proton specific channel.
In linear or non cyclic electron transport of photosynthesis NADPH a.pdf
In linear or non cyclic electron transport of photosynthesis NADPH and ATP are produced.
These molecules are then used for reduction during dark reaction.
Solution
In linear or non cyclic electron transport of photosynthesis NADPH and ATP are produced.
These molecules are then used for reduction during dark reaction..
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Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
PART A The element Si belongs to IVA group. Therefore, four electro.pdf
- 1. PART A: The element Si belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D: All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons. PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is [Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal. PART I: The element Ba belongs to IIA group. Therefore, two electrons present in the s-orbital. Hence the condensed electron configuration of Ba is [Xe] 6s2. PART J: The outermost s- electron is completely filled. Therefore, there are no unpaired electrons present in Ba metal. PART K: The element Tl belongs to IIIA group. Therefore, the condensed electron configuration of Tl is [Xe] 4f14 5d10 6s2 6p1. PART L: The all orbitals are filled with electrons except Tl. Therefore, the number of unpaired electrons is 1 Solution PART A: The element Si belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Si is [Ne] 3s23p2. PART B: The s orbital is completely filled and there are two electrons present in p-orbital. Therefore, the number of unpaired electrons of Si is 2 PART C: The atomic number of Zn is 30 and it belongs to IIB group. Therefore, the condensed electron configuration of Zn is [Ar] 4s23d10. PART D: All orbitals of Zn completely filled with electrons. Therefore, there are no unpaired electrons. PART E: The atomic number of Zr is 40 and it belongs to IVB group. This element is present after third element of Kr in periodic table. Hence the condensed electron configuration of Zr is [Kr] 5s24d2. PART F: The s orbital is completely with electrons and there are two electrons present in the d orbital. Therefore, the number of unpaired electrons of Zr is 2 PART G: The element Sn belongs to IVA group. Therefore, four electrons present in the outermost shell. Hence the condensed electron configuration of Sn is [Kr] 5s24d10 5p2. PART H: The s orbital and p orbital is completely filled. Therefore, there are two unpaired electrons in the Sn metal. PART I: The element Ba belongs to IIA group. Therefore, two electrons present in the s-orbital. Hence the condensed electron configuration of Ba is [Xe] 6s2. PART J: The outermost s-
- 2. electron is completely filled. Therefore, there are no unpaired electrons present in Ba metal. PART K: The element Tl belongs to IIIA group. Therefore, the condensed electron configuration of Tl is [Xe] 4f14 5d10 6s2 6p1. PART L: The all orbitals are filled with electrons except Tl. Therefore, the number of unpaired electrons is 1